Variety of solutions depending on constraints

Revision en3, by Coder, 2015-09-13 21:33:59

Suppose the input for a problem is positive integer N and answer happens to be 1/N. Here is what happens for different constraints on N. You decide which version is the hardest.

1.

N <= 10^9

Everyone does math and all solutions look like

cout << 1.0 / N

2.

N <= 10^6

You see solutions from above but some contestants decide to use iterative approach, something like

double ans = 1.0;
for (int i = 2; i <= N; i++) ans = ans * (i - 1) / i;

3.

N <= 10^5

Mostly solutions from above, but some contestants use recursion (maybe with memoization)

double f(int N) {
   if (N == 1) return 1.0;
   return f(N - 1) * (N - 1) / N;
}

4.

N <= 500

Solutions from above, but some contestants go with O(N^3) dynamic programming instead

5.

N <= 109

Contestants think there is a typo and go with first solution

6.

N <= 100

Some contestants realize that you can precompute all answers

const double ans[100]={1.0, 0.5, 0.333333333, ... }

7.

N <= 47

Intended solution from Ukrainian author has O(N^4) complexity

8.

N <= 36

You see lot of solutions with meet-in-the middle approach

9.

N <= 20

You start to see a lot of bitwise operations in codes that can mean dynamic programming on subsets

10.

N <= 10

Some C++ codes use next_permutation(), Java coders having harder time

History

 
 
 
 
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  Rev. Lang. By When Δ Comment
en3 English Coder 2015-09-13 21:33:59 63 Tiny change: 'n\n\n\n\n---\nNot sure which version is the hardest. Any ideas?' - (published)
en2 English Coder 2015-09-13 21:28:00 486 Tiny change: ' '
en1 English Coder 2015-09-13 21:08:03 1147 Initial revision (saved to drafts)